Lemma 51.20.1. Let $R$ be a ring. Let $M \to M'$ be a map of $R$-modules with $M$ of finite presentation such that $\text{Tor}_1^ R(M, N) \to \text{Tor}_1^ R(M', N)$ is zero for all $R$-modules $N$. Then $M \to M'$ factors through a free $R$-module.

## 51.20 A bit of uniformity, I

The main task of this section is to formulate and prove Lemma 51.20.2.

**Proof.**
We may choose a map of short exact sequences

whose right vertical arrow is the given map. We can factor this map through the short exact sequence

which is the pushout of the first short exact sequence by $K \to K'$. By a diagram chase we see that the assumption in the lemma implies that the boundary map $\text{Tor}_1^ R(M, N) \to K' \otimes _ R N$ induced by (51.20.1.1) is zero, i.e., the sequence (51.20.1.1) is universally exact. This implies by Algebra, Lemma 10.82.4 that (51.20.1.1) is split (this is where we use that $M$ is of finite presentation). Hence the map $M \to M'$ factors through $\bigoplus _{i \in I} R$ and we win. $\square$

Lemma 51.20.2. Let $R$ be a ring. Let $\alpha : M \to M'$ be a map of $R$-modules. Let $P_\bullet \to M$ and $P'_\bullet \to M'$ be resolutions by projective $R$-modules. Let $e \geq 0$ be an integer. Consider the following conditions

We can find a map of complexes $a_\bullet : P_\bullet \to P'_\bullet $ inducing $\alpha $ on cohomology with $a_ i = 0$ for $i > e$.

We can find a map of complexes $a_\bullet : P_\bullet \to P'_\bullet $ inducing $\alpha $ on cohomology with $a_{e + 1} = 0$.

The map $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is zero for all $R$-modules $N$ and $i > e$.

The map $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N)$ is zero for all $R$-modules $N$.

Let $N = \mathop{\mathrm{Im}}(P'_{e + 1} \to P'_ e)$ and denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N)$ the canonical element (see proof). Then $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N)$.

The map $\text{Tor}_ i^ R(M, N) \to \text{Tor}_ i^ R(M', N)$ is zero for all $R$-modules $N$ and $i > e$.

The map $\text{Tor}_{e + 1}^ R(M, N) \to \text{Tor}_{e + 1}^ R(M', N)$ is zero for all $R$-modules $N$.

Then we always have the implications

If $M$ is $(-e - 1)$-pseudo-coherent (for example if $R$ is Noetherian and $M$ is a finite $R$-module), then all conditions are equivalent.

**Proof.**
It is clear that (2) implies (1). If $a_\bullet $ is as in (1), then we can consider the map of complexes $a'_\bullet : P_\bullet \to P'_\bullet $ with $a'_ i = a_ i$ for $i \leq e + 1$ and $a'_ i = 0$ for $i \geq e + 1$ to get a map of complexes as in (2). Thus (1) is equivalent to (2).

By the construction of the $\mathop{\mathrm{Ext}}\nolimits $ and $\text{Tor}$ functors using resolutions (Algebra, Sections 10.71 and 10.75) we see that (1) and (2) imply all of the other conditions.

It is clear that (3) implies (4) implies (5). Let $N$ be as in (5). The canonical map $\tilde\xi : P'_{e + 1} \to N$ precomposed with $P'_{e + 2} \to P'_{e + 1}$ is zero. Hence we may consider the class $\xi $ of $\tilde\xi $ in

Choose a map of complexes $a_\bullet : P_\bullet \to P'_\bullet $ lifting $\alpha $, see Derived Categories, Lemma 13.19.6. If $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N)$, then we find a map $\varphi : P_ e \to N$ such that $\tilde\xi \circ a_{e + 1} = \varphi \circ d$. Thus we obtain a map of complexes

as in (2). Hence (1) – (5) are equivalent.

The equivalence of (6) and (7) follows from dimension shifting; we omit the details.

Assume $M$ is $(-e - 1)$-pseudo-coherent. (The parenthetical statement in the lemma follows from More on Algebra, Lemma 15.64.17.) We will show that (7) implies (4) which finishes the proof. We will use induction on $e$. The base case is $e = 0$. Then $M$ is of finite presentation by More on Algebra, Lemma 15.64.4 and we conclude from Lemma 51.20.1 that $M \to M'$ factors through a free module. Of course if $M \to M'$ factors through a free module, then $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is zero for all $i > 0$ as desired. Assume $e > 0$. We may choose a map of short exact sequences

whose right vertical arrow is the given map. We obtain $\text{Tor}_{i + 1}^ R(M, N) = \text{Tor}^ R_ i(K, N)$ and $\mathop{\mathrm{Ext}}\nolimits ^{i + 1}_ R(M, N) = \mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N)$ for $i \geq 1$ and all $R$-modules $N$ and similarly for $M', K'$. Hence we see that $\text{Tor}_ e^ R(K, N) \to \text{Tor}_ e^ R(K', N)$ is zero for all $R$-modules $N$. By More on Algebra, Lemma 15.64.2 we see that $K$ is $(-e)$-pseudo-coherent. By induction we conclude that $\mathop{\mathrm{Ext}}\nolimits ^ e(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^ e(K, N)$ is zero for all $R$-modules $N$, which gives what we want. $\square$

Lemma 51.20.3. Let $I$ be an ideal of a Noetherian ring $A$. For all $n \geq 1$ there exists an $m > n$ such that the map $A/I^ m \to A/I^ n$ satisfies the equivalent conditions of Lemma 51.20.2 with $e = \text{cd}(A, I)$.

**Proof.**
Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ n, N)$ be the element constructed in Lemma 51.20.2 part (5). Since $e = \text{cd}(A, I)$ we have $0 = H^{e + 1}_ Z(N) = H^{e + 1}_ I(N) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{e + 1}(A/I^ m, N)$ by Dualizing Complexes, Lemmas 47.10.1 and 47.8.2. Thus we may pick $m \geq n$ such that $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ m, N)$ as desired.
$\square$

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